Problem: Find $\cos\left(\dfrac{19\pi}{12}\right)$ exactly using an angle addition or subtraction formula. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-\sqrt{8}}{3}$ (Choice B) B $\dfrac{-1-\sqrt{3}}{2}$ (Choice C) C $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (Choice D) D $\dfrac{1+\sqrt{3}}{2}$
Answer: The strategy First, we should rewrite the given angle $\dfrac{19\pi}{12}$ as the sum or difference of two special angles. Then, we can use the cosine addition or subtraction identities in order to evaluate $\cos\left(\dfrac{19\pi}{12}\right)$. [How do we find the trigonometric value of a sum or difference?] Rewriting $\dfrac{19\pi}{12}$ We can rewrite $\dfrac{19\pi}{12}$ as follows. $\begin{aligned}\dfrac{19\pi}{12}&=\dfrac{15\pi}{12}+\dfrac{4\pi}{12}\\\\\\ &=\dfrac{5\pi}{4}+\dfrac{\pi}{3}\end{aligned}$ In other words, $\dfrac{19\pi}{12}$ is the sum of the special angles $\dfrac{5\pi}{4}$ and $\dfrac{\pi}{3}$. Evaluating $\cos\left(\dfrac{19\pi}{12}\right)$ Using the cosine addition identity, we get the following. $\begin{aligned} \cos\left(\dfrac{19\pi}{12}\right)&= \cos\left(\dfrac{5\pi}{4}+\dfrac{\pi}{3}\right) \\\\\\ &= \cos \left(\dfrac{5\pi}{4}\right) \cos \left(\dfrac{\pi}{3}\right) - \sin \left(\dfrac{5\pi}{4}\right) \sin \left(\dfrac{\pi}{3}\right) \\\\\\ &=\left(-\dfrac{\sqrt{2}}{2}\right) \left(\dfrac{1}{2}\right) - \left(-\dfrac{\sqrt{2}}{2}\right) \left(\dfrac{\sqrt{3}}{2}\right) \\\\\\ &=\left(-\dfrac{\sqrt{2}}{4}\right) + \left(\dfrac{\sqrt{6}}{4}\right)\\\\\\ &=\dfrac{\sqrt{6}-\sqrt{2}}{4} \end{aligned}$ Summary $\cos\left(\dfrac{19\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}$